Thermal Energy I don't see why my method is wrong.

[Null_]

Homework Statement

300 g of water whose temp is 25*C are added to a thin glass containing 800 g of water at 20*C. What is the final temperature of the water?

Homework Equations

deltaE=mCdeltaT

The Attempt at a Solution

Here's what I did:
deltaEsys=mCdeltaT
[300g (4.2 J/K/g) (Tf - 298K) ] - [800g (4.2J/K/g)(Tf- 293K) ] = 0
(300g)(Tf-298K) = (800g)(Tf-293K)
300g*Tf - 89400g*K = 800g*Tf - 234400g*K
145000g*K = 500 g*Tf
290K = Tf

The correct answer is 21.36*C (294.36 K)

 

[NobodySpecial]

If you are going to use different signs for the temperature difference then you need to add the two energy changes.

 

[ehild]

[300g (4.2 J/K/g) (Tf - 298K) ] - [800g (4.2J/K/g)(Tf- 293K) ] = 0

that "-" should be "+"

ehild

 

[Null_]

Ah, thank you. I was thinking about: Ef-Ei=W=0, but that is obviously not the case since there are two initials and two finals.

 

[tiny-tim]

Hi Null_!

(have a degree: ° )

300 g of water whose temp is 25*C are added to a thin glass containing 800 g of water at 20*C. What is the final temperature of the water?
…
[300g (4.2 J/K/g) (Tf - 298K) ] - [800g (4.2J/K/g)(Tf- 293K) ] = 0
(300g)(Tf-298K) = (800g)(Tf-293K)
300g*Tf - 89400g*K = 800g*Tf - 234400g*K
145000g*K = 500 g*Tf
290K = Tf

The correct answer is 21.36*C (294.36 K)

Why did you use the 4.2 and the 293 ?

This is a straightforward averaging question … no point in multiplying by, or subtracting, constants that have to be eliminated later …

no wonder you made a mistake somewhere!

try it again, just finding the average temperature when 3/11 is at 25° and 8/11 is at 20° ​

 

[NobodySpecial]

Ah, thank you. I was thinking about: Ef-Ei=W=0,

That's correct but by making one of the dT negative you did Ef - (-Ei) = 0

 

[Dickfore]

Don't listen to 'tiny-tim'. While his method works in this particular case, it should be derived from the law of conservation of the total internal energy of the system (since the system is thermally isolated and there is no work done on it).

As a reminder that 'tiny-tim''s rule is incorrect in general, how would you calculate the equilibrium temperature if you had some quantity of ice at -5 ^oC and you pour over some quantity of hot water at 80 ^o?

 

[tiny-tim]

As a reminder that 'tiny-tim''s rule is incorrect in general, how would you calculate the equilibrium temperature if you had some quantity of ice at -5 ^oC and you pour over some quantity of hot water at 80 ^o?

Ah, but that's two different materials (or two different states, anyway) …

here, it's the same state of the same material, so the averaging method is correct (technically as well as practically).

 

[Dickfore]

Ah, but that's two different materials (or two different states, anyway) …

here, it's the same state of the same material, so the averaging method is correct (technically as well as practically).

What if the heat capacity is temperature dependent?

 

[tiny-tim]

What if the heat capacity is temperature dependent?

I've never come across a question involving that!